6

I would like someone to verify the calcuations that I'm making about the tangent spaces of $O(n)$, and help answer some general questions that I have from this special case. I don't know much about Lie theory at all, so I'm flying by the seat of my pants here :)

So, we have the space $O(n) \equiv \{ X | X^TX = I \}$. It's not hard to show that the tangent space should satisfy the constraint: tangent space at $P \in O(n) \equiv T_P O(n) \equiv \{ Z | Z^T P + P^T Z = 0\}$.

Now, by performing the usual Lie algebra trick of "calculate tangent space at identity", we see that:

$$ T_I O(n) = \{ Z | Z^T + Z = 0 \} $$

Next, to calculate the tangent space at some arbitrary point $P \in O(n)$, we consider the map: $$ f: O(n) \rightarrow O(n) \qquad f(X) = PX $$ Note that $f(I) = P$, and hence the differential of this map, $df$ will have the type $df: T_I O(n) \rightarrow T_P O(n)$. Now, all that's left to do is to calculate the differential.

To perform this, we take a curve: $$ c: (-\epsilon, \epsilon) \rightarrow O(n) \qquad c(t) = e^{Kt} \qquad K^T = -K $$

This has image in $O(n)$, since

$$ c(t)^T c(t) = e^{K^T t}e^{Kt} = e^{t(K^T + K)} = e^{t\cdot 0} = I $$

Hence, this is a valid curve.

Now, we compute $df$:

$$ df \equiv \frac{d}{dt} (f \circ c)(t) \vert_{t = 0} = \frac{d}{dt} P e^{Kt} \vert_{t=0} = PKe^{K0} = PK $$

Hence, the tangent space at the point $P$ is $T_PO(n) \equiv \{ PK | K^T + K = 0 \}$

I am uncomfortable with many things that "automatically work" in this proof:

  1. Is this proof correct? If not, where is it wrong?

  2. Why is the choice of the integral curve $c(t) = e^{Kt}$ the "correct" choice? How do I prove that the mapping of tangent spaces is indeed a bijection, and not some artefact of the curve parametrization I chose?

  3. What is the "general form" of this proof, for an arbitrary matrix Lie group $M$? Can I state that the tangent space at a point $P$ will be of the form $T_P M \equiv \{ PZ \mid Z \in T_I M \}$? If not, why is this not correct?

  4. Is there a slick proof of this fact?

Rodrigo de Azevedo
  • 1
Siddharth Bhat
  • 8,452

1 Answers1

2
  1. Yes, this proof is correct.
  2. You could use any curve $c:(-\epsilon, \epsilon)\to O(n)$ that satisfies $c(0) = I$ and $c'(0)=K$. Your calculation of $"df"$ (actually, you should write $(df)_I(K)$) would be identical.
  3. For a matrix Lie group $M$, this is pretty much the general form of the proof. For a general Lie group (which does not necessarily have a matrix representation), the only difference would be that since $f(X) = L_PX$, where $L_P:G\to G$ denotes left multiplication by $P\in G$, you would have $$ (df)_I(K) = (dL_P)_I(K). $$ This expression is commonly just written as $P\cdot K$ for $P\in G, K\in\mathfrak{g}$, however.
  4. Not really, as far as I can tell. Maybe you'd prefer this one: let $d:(-\epsilon,\epsilon)\to G$ be an arbitrary curve through $P\in G$. Then $t\mapsto P^{-1}\cdot d(t)$ is a curve passing through $I$, so its derivative $P^{-1}\cdot d'(0)\in T_IG = \mathfrak{g}$. Hence $d'(0)\in P\cdot\mathfrak{g}$, implying $T_PG = P\cdot\mathfrak{g}$. Incidentally, using $d(t)\cdot P^{-1}$ instead would yield $T_PG = \mathfrak{g}\cdot P$, which is also true. This proof is essentially the same as the one you gave, however.
user17945
  • 3,943